solution38

# solution38 - cherubin(mcc2645 – Quizzes 2 Electric Field...

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Unformatted text preview: cherubin (mcc2645) – Quizzes 2 Electric Field and Potential – Catala – (12054) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two point charges of magnitude 5 . 74 nC and- 4 . 04 nC are separated by 43 . 7 cm. The acceleration of gravity is 9 . 8 m / s 2 . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . What is the electric potential at a point midway between the charges? Correct answer: 69 . 926 V. Explanation: Let : q 1 = 5 . 74 nC = 5 . 74 × 10- 9 C , q 2 =- 4 . 04 nC =- 4 . 04 × 10- 9 C , r = 43 . 7 cm = 0 . 437 m , r 1 = r 2 = r 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The electric potential is V = V 1 + V 2 = k e r 1 q 1 + k e r 2 q 2 = 2 k e r ( q 1 + q 2 ) = 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) . 437 m × bracketleftbig (5 . 74 × 10- 9 C) + (- 4 . 04 × 10- 9 C) bracketrightbig = 69 . 926 V ....
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## This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

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solution38 - cherubin(mcc2645 – Quizzes 2 Electric Field...

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